Download 3 - Utrecht University Repository

Power Values of Divisor Sums
Author(s): Frits Beukers, Florian Luca, Frans Oort
Reviewed work(s):
Source: The American Mathematical Monthly, Vol. 119, No. 5 (May 2012), pp. 373-380
Published by: Mathematical Association of America
Stable URL: http://www.jstor.org/stable/10.4169/amer.math.monthly.119.05.373 .
Accessed: 15/02/2013 04:05
Your use of the JSTOR archive indicates your acceptance of the Terms & Conditions of Use, available at .
http://www.jstor.org/page/info/about/policies/terms.jsp
.
JSTOR is a not-for-profit service that helps scholars, researchers, and students discover, use, and build upon a wide range of
content in a trusted digital archive. We use information technology and tools to increase productivity and facilitate new forms
of scholarship. For more information about JSTOR, please contact [email protected].
.
Mathematical Association of America is collaborating with JSTOR to digitize, preserve and extend access to
The American Mathematical Monthly.
http://www.jstor.org
This content downloaded on Fri, 15 Feb 2013 04:05:44 AM
All use subject to JSTOR Terms and Conditions
Power Values of Divisor Sums
Frits Beukers, Florian Luca, and Frans Oort
Abstract. We consider positive integers whose sum of divisors is a perfect power. This problem had already caught the interest of mathematicians from the 17th century like Fermat,
Wallis, and Frenicle. In this article we study this problem and some variations. We also give
an example of a cube, larger than one, whose sum of divisors is again a cube.
1. INTRODUCTION. Recently, one of the current authors gave a mathematics
course for an audience with a general background and age over 50. Although some
non-mathematicians have developed a fear and a dislike for mathematics, the course
was well attended, and the participants eagerly followed many aspects of pure mathematics. At one occasion, during the discussion of divisor sums, one participant asked:
“the sum of the divisors of my age is a square, can you tell me how old I am?” She did
not look like she was of age 1 or 3, or even 22 for that matter, so the teacher answered;
“Well I cannot tell you exactly, but I do know your age must be one of 66, 70, 81 or
94”. The teacher then asked; “do you think there are infinitely many integers of which
the divisor sum is a square?” This was a nice example of the theme of the course, how
do mathematicians arrive at their conjectures? Moreover, how do interesting questions
stimulate the search for new mathematics? The teacher’s question is also the starting
theme of this note.
2. THE PROBLEM. By the divisor sum of an integer n, we mean the sum of all divisors, including n itself, and we denote it by σ (n). Divisor sums are among the oldest
quantities studied in number theory. Already in Greek times people were interested in
perfect numbers. That is, a number n whose sum of divisors, excluding n itself, equals
n again. In terms of the σ -function this comes down to σ (n) = 2n. An important feature of σ (n) is its multiplicative property. This means that if m and n are co-prime
(greatest common divisor one) then σ (mn) = σ (m)σ (n). This reduces the computation of σ (n) to a factorization of n and a computation of values of divisor sums of
primes or prime powers. Clearly, for a prime p we have σ ( p) = p + 1 and for its
kth power σ ( p k ) = 1 + p + · · · + p k . For a discussion and proof of the multiplicative
property of σ one can consult most books in elementary number theory, for example
[7, Section 16.7]. The following theorem answers the teacher’s question.
Theorem 2.1. There exist infinitely many positive integers n such that σ (n) is the
square of an integer.
We cannot attach the name of a first discoverer of this theorem, but it is very likely
that this theorem has been proven many times in the history of number theory. The
proof is not hard and will be given in Section 3, but we can already explain its main
idea here. We write down the factorizations of σ ( p) for all primes p ≤ 23:
σ (2) = 3,
σ (3) = 22 ,
http://dx.doi.org/10.4169/amer.math.monthly.119.05.373
MSC: Primary 11N25, Secondary 11N64
May 2012]
POWER VALUES OF DIVISOR SUMS
This content downloaded on Fri, 15 Feb 2013 04:05:44 AM
All use subject to JSTOR Terms and Conditions
373
σ (5) = 2 × 3,
σ (7) = 23 ,
σ (11) = 22 × 3,
σ (13) = 2 × 7,
σ (17) = 2 × 32 ,
σ (19) = 22 × 5,
σ (23) = 23 × 3.
There are plenty of ways in which we can make a selection of these equalities such that
after multiplication the right hand side is a square. For example, multiply σ (2) = 3
and σ (11) = 22 × 3 to get σ (2)σ (11) = 22 × 32 . By the multiplicative property of
σ we get σ (22) = 22 × 32 . Similarly, the factorizations of σ (7) and σ (17) give us
σ (7 × 17) = 24 × 32 , again a square. To find more of such relations we symbolize a
factorization 2a 3b 5c 7d by its so-called exponent vector (a, b, c, d). We then need to
find a subset of these vectors whose sum is (0, 0, 0, 0) modulo 2. In other words, we
look for linear relations between the exponent vectors modulo 2. Our problem is thus
reduced to linear algebra over the field of two elements, F2 . In the example above, the
rank of the 9 exponent vectors over F2 is 4. Hence the space of relations has dimension
9 − 4 = 5 and thus we can construct 25 − 1 = 31 non-trivial relations in this way. This
gives us 31 examples of solutions m, n to σ (n) = m 2 . It turns out that by increasing
the number of factorizations, the number of values n for which σ (n) will increase very
quickly.
To continue the teacher’s investigations, it is a pleasant surprise to find that there
exist integers whose divisor sum is a cube, or a fourth power, or higher powers. The
smallest examples of σ (n) = m 3 with n > 1 are σ (7) = 23 , σ (102) = 63 , . . . and the
smallest fourth power example is σ (217) = 44 (incidently, 44 = 28 ). We have the following theorem.
Theorem 2.2. Given a positive integer k, there exist infinitely many positive integers
n such that σ (n) is the kth power of an integer.
Although we found a proof of this theorem in the beginning of 2010, the ingredients
have been around for a much longer time. So there is a fair chance that it has been observed and proven before. We will give a proof in Section 4. However, there is a much
more surprising result. It can be found in a recent preprint by T. Freiberg [5], which
deals with the problem of estimating the number of kth powers in products of shifted
primes. For the particular case of divisor sums it has the following consequence.
Theorem 2.3 (Freiberg, 2010). Let k be a positive integer and let sk (x) be the number
of positive integers n < x for which σ (n) is a kth power. Then there exists a positive
real number xk such that
sk (x) > x 0.7
for all x ≥ xk .
Naively one might have guessed that sk (x) would be about x 1/k , comparable to the
number of kth powers below x. So this theorem is indeed remarkable. In a slightly
374
c THE MATHEMATICAL ASSOCIATION OF AMERICA [Monthly 119
This content downloaded on Fri, 15 Feb 2013 04:05:44 AM
All use subject to JSTOR Terms and Conditions
exaggerated way one might say that the values of σ are really eager to be perfect
powers. The proof can be found in [5, Theorem 1.1], where an exponent slightly larger
than 0.7 is given. We are thankful to the referee for pointing out Freiberg’s preprint to
us. The paper [2] contains the ingredients for the proof of this theorem. More precisely,
[2] deals with square values of the Euler totient function φ, but similar arguments apply
to σ .
The idea behind the proof of the two theorems above is again to write down factorizations of many σ ( p), with p prime, and choose factorizations from this list
whose product is a kth power. In particular take k = 3 and the factorizations of
σ (2), . . . , σ (23) given above. This time the use of linear algebra in finding linear relations modulo 3 between the exponent vectors is not going to help us. We are interested
only in relations with coefficients 0, 1. Straightforward solution of linear equations
modulo 3 would also give us coefficients 2, which are not useful to us. The following
theorem comes to our rescue.
Theorem 2.4. Let k be a prime power and r a positive integer. Then any subset of
the additive group (Z/kZ)r with more than r (k − 1) elements contains a non-empty
subset whose sum of elements is zero.
This theorem, in the language of abelian groups, was proved independently by
Van Emde Boas, Kruyswijk [4] and Olson [10]. Unfortunately [4] is difficult to come
by, but in [10] one finds a simple and delightful proof. Application in the case k = 3
shows that if the number of factorizations exceeds twice the length of the exponent vectors, then there is a relation. In our example there are 8 factorizations with exponent
vectors of length 3. So we should be able to find relations. For example, σ (7) = 23 ,
σ (2 × 5 × 11) = 23 × 33 and σ (3 × 17 × 23) = 26 × 33 . A computer search shows
that we can construct 15 examples of σ (n) = m 3 in this way. We used a computer
search this time, because Theorem 2.4 is only an existence theorem. The actual construction of a subset with sum zero can be a very difficult problem, especially for larger
r , say r > 100.
In Neil Sloane’s online encyclopedia of integer sequences [11] we find several sequences dedicated to this problem, namely A006532 (σ (n) = m 2 ), A020477 (σ (n) =
m 3 ), A019422 (σ (n) = m 4 ), A019423 (σ (n) = m 5 ), A019424 (σ (n) = m 6 ), A048257
(σ (n) = m 7 ) and A048258 (σ (n) = m 8 ).
To top off our story, we turn to a very general question.
Question 2.5. Given positive integers l ≥ 1 and k ≥ 2, do there exist integers n such
that σ (n l ) is a kth power?
In Dickson’s History of the Theory of Numbers [3, Vol I, Chapter 2, p. 54] we find
that Fermat observed σ (73 ) = 202 and asked for more examples of σ (n 3 ) = m 2 and,
as second question, σ (n 2 ) = m 3 . Subsequently, Wallis, Frenicle and many others came
up with additional examples. Their strategy is similar to the one we used above. Write
down the prime factorizations of σ ( p 3 ) for all primes p up to a certain limit. By way of
example we record the prime factorizations of σ ( p 3 ) for all p < 50 and prime factors
less than 50:
σ 23 = 3 × 5,
σ 33 = 23 × 5,
σ 53 = 22 × 3 × 13,
May 2012]
POWER VALUES OF DIVISOR SUMS
This content downloaded on Fri, 15 Feb 2013 04:05:44 AM
All use subject to JSTOR Terms and Conditions
375
σ 73 = 24 × 52 ,
σ 133 = 22 × 5 × 7 × 17,
σ 173 = 22 × 32 × 5 × 29,
σ 313 = 26 × 13 × 37,
σ 413 = 22 × 3 × 7 × 292 ,
σ 433 = 23 × 52 × 11 × 37,
σ 473 = 25 × 3 × 5 × 13 × 17.
The fact that σ ( p 3 ) = ( p + 1)( p 2 + 1) certainly helps in getting small primes in the
factorizations. We have listed 10 factorizations with 9 distinct primes occurring. Hence
we can find a subset of factorizations whose product is a square. For example,
σ (2 × 3 × 5 × 13 × 41 × 47)3 = 214 × 34 × 54 × 72 × 132 × 172 × 292 .
This, by the way, is the smallest example of σ (n 3 ) = m 2 beyond σ (13 ) = 12 and Fermat’s σ (73 ) = 202 . It was found by Wallis in 1658 (see [3, Vol I, p. 55]). Increasing
the number of primes, and the bound on the allowed prime factors, we find many
more such possibilities. For example, listing the factorization of σ ( p 3 ) for all primes
p < 10000 and prime factors < 2000 gives us 251 factorizations with 182 distinct
primes. This provides us with at least 2251−182 − 1 = 269 − 1 examples of σ (n 3 ) = m 2 .
As for Fermat’s second question, σ (n 2 ) = m 3 , the smallest example (after σ (12 ) =
3
1 ) is
σ (2 × 3 × 11 × 653)2 = 17293 ,
found relatively late by A.S. Bang in 1878 (see [3, Vol I, p. 56]). This belated discovery
has probably something to do with the large prime 653. Several larger examples had
been found earlier by Wallis and Frenicle (see [3, Vol I, pp. 55, 56]). To find such
examples, one can write down all factorizations of σ ( p 2 ) for the first, say K , primes
p and whose prime factors are less than a relatively small bound, say B. For example,
taking K = 1000 and B = 100 we find 26 factorizations with 12 distinct primes. Since
26 is larger than twice 12, an application of Theorem 2.4 shows that there exists a
subset of these factorizations whose product is a cube.
But there is more. Additional computer calculations of a similar kind seem to suggest the following two conjectures.
Conjecture 2.6. For every positive integer k there exist infinitely many positive integers n such that σ (n 2 ) is kth power of an integer.
Conjecture 2.7. For every positive integer k there exist infinitely many positive integers n such that σ (n 3 ) is kth power of an integer.
For example, we found that among the factorizations of the first 10000 primes p there
are 280 factorizations of σ ( p 3 ) with prime factors < 1000. The number of distinct
prime factors that occur is 125. So there are more than twice as many factorizations
as there are prime factors. Hence, Theorem 2.4 tells us there exists an example of
σ (n 3 ) = m 3 . However, finding it is quite a computational challenge. Herman te Riele
was kind enough to make such an attempt on our behalf and reported the following in a
376
c THE MATHEMATICAL ASSOCIATION OF AMERICA [Monthly 119
This content downloaded on Fri, 15 Feb 2013 04:05:44 AM
All use subject to JSTOR Terms and Conditions
personal communication. He considered 63 factorizations of σ ( p 3 ) with prime factors
< 200. The number of primes that occur in the factorizations is 46. Although Theorem
2.4 does not apply here, Te Riele found nevertheless that σ (n 3 ) = m 3 with
n = 5 × 11 × 13 × 17 × 41 × 47 × 193 × 239 × 443 × 499
× 701 × 1087 × 3583 × 5507 and
m = 217 × 35 × 56 × 7 × 134 × 172 × 29 × 37 × 61 × 97 × 149 × 157.
It would be interesting to know if this is the smallest solution of σ (n 3 ) = m 3 with
n > 1. We were unable to find very much on the problem σ (n l ) = m k with k, l ≥ 2 in
modern literature. In Neil Sloane’s online encyclopedia of integer sequences [11] we
find several sequences dedicated to this problem, A008847 (σ (n 2 ) = m 2 ), A008850
(σ (n 2 ) = m 3 ) and A008849 (σ (n 3 ) = m 2 ).
For any pair (l, k) with l > 3, k > 1 we expect that the number of solutions to
σ (n l ) = m k is finite.
3. THE CASE OF SQUARE VALUES. We give an elementary proof of Theorem
2.1 which is based on the ideas explained in the previous section. We have seen that
finding examples of σ (n) = m 2 is not hard. The only concern is that we like to be
certain to find infinitely many such examples. The proof we present here works from
scratch and doesn’t use any results on the distribution of prime numbers. It is based on
an idea of Ronald van Luijk, published anonymously in [9].
Let us denote the sequence of primes by p1 = 2, p2 = 3, . . . . Choose a positive
r
integer N > 2. For every i let ri be the smallest integer such that pi i > NQ
. Clearly ri =
r
1 if pi > N . Choose t such that pt is larger than every prime factor in pi ≤N σ ( pi i ).
For any pi > N we have σ ( pi ) = 2( pi + 1)/2. Hence, the prime factors of σ ( pi ) are
r
strictly less than pi . Thus we see that every factorization of σ ( pi i ) with i ≤ t consists
of primes strictly less than pt . We write down the factorizations
ai,t−1
a
a
r σ pi i = p1i1 p2i2 · · · pt−1
, for i = 1, . . . , t.
The number of prime factorizations is t and the total number of prime factors in these
factorizations is strictly less than or equal to t − 1. So in the t × (t − 1)-matrix (ai j ),
the number of rows exceeds the length of the rows. Hence, it follows from linear
algebra over the field F2 of two elements that we can find i ∈ {0, 1} with 1 ≤ i ≤ t,
not all zero, such that
1 (a11 , . . . , a1,t−1 ) + · · · + t (at1 , . . . , at,t−1 )
has even entries. Hence,
r
σ p11
t
· · · σ ptrt
t
r
= σ p11 1 · · · ptt rt
is a square. Note that we have constructed an example of σ (n) = m 2 where the prime
power factors of n are all larger than N . By choosing N as big as we like we can
construct infinitely many such examples. This proves Theorem 2.1.
4. HIGHER POWERS. As a curiosity we mention the following.
Remark 4.1. Let k be any integer ≥ 2. If there are infinitely many Mersenne primes
(i.e., primes of the form 2 p − 1) then there are infinitely many integers n such that
σ (n) is the kth power of an integer.
May 2012]
POWER VALUES OF DIVISOR SUMS
This content downloaded on Fri, 15 Feb 2013 04:05:44 AM
All use subject to JSTOR Terms and Conditions
377
Proof. Denote the set of exponents { p : 2 p − 1 prime} by P . We have assumed that
this set is infinite. If we consider the elements of P modulo k, then there is a residue
class a (mod k) which contains infinitely many elements of P . Choose k elements
p1 , . . . , pk ∈ P such that p1 ≡ · · · ≡ pk ≡ a (mod k). Then clearly, p1 + · · · + pk ≡
ka ≡ 0 (mod k). Observe that
σ (2 p1 − 1) × · · · × (2 pk − 1) = 2 p1 × · · · × 2 pk .
The latter number equals 2 p1 +···+ pk whose exponent we have seen to be divisible by
k. Hence we have a kth power value of σ . Since there are infinitely many choices of
p1 , . . . , pk , we get infinitely kth power values of σ . This proves the remark.
However, it is unknown whether there are infinitely many Mersenne primes. So
we had better look for another proof of Theorem 2.2. The following proof uses more
advanced methods than in the case of square values. One of the ingredients is the prime
number theorem.
Theorem 4.2. If π(x) is the prime counting function, then
π(x) ∼
x
log x
as x → ∞.
The ∼ sign means that the ratio of the functions tends to 1 as x → ∞. For a proof,
see [12]. In addition we need the following.
Proposition 4.3. If M > 1 is given, then there exists C > 0 such that the number of
primes p ≤ x, where p + 1 is divisible by a prime ≥ p/M, is bounded from above by
C x/(log x)2 ) as x → ∞.
Proof. The proposition depends on a sieve result which can be found in the book by
Halberstam and Richert [6, Theorem 3.12]. This theorem implies that, given an even
integer a, the number of primes q < x, such that aq − 1 is also prime, is bounded
from above by a constant Ca times x/(log x)2 .
To apply this to our situation, let p with M < p ≤ x be a prime such that σ ( p) has
a prime divisor ≥ p/M. This means that there exists a prime q and an even integer a <
M + 1 such that p + 1 = aq. Certainly q < x. From the sieve
P result, it follows that
the number of such pairs (a, q) is bounded from above by ( a<M+1 Ca )x/(log x)2 .
Hence, the number of primes p ≤ x such that p + 1 is divisible by a prime ≥ p/M is
bounded from above by a constant times x/(log x)2 .
Proposition 4.3 tells us that the proportion of the primes p ≤ x such that p + 1 is
divisible by a prime ≥ p/M is bounded from above by C x/(π(x)(log x)2 ). So this
proportion is bounded from above by C/ log x, which goes to 0 when we let x → ∞.
Hence, the number of primes p ≤ x whose prime factorization does not contain a
prime factor ≥ p/M is still ∼ π(x) ∼ x/ log x. This is what we will exploit in the
following proof.
Proof of Theorem 2.2. Choose M > k log k and choose
√ x sufficiently large. Write
down the factorizations of σ ( p) for all primes p with x ≤ p ≤ x. Then delete all
factorizations which contain a prime factor ≥ p/M. According to Proposition 4.3 this
378
c THE MATHEMATICAL ASSOCIATION OF AMERICA [Monthly 119
This content downloaded on Fri, 15 Feb 2013 04:05:44 AM
All use subject to JSTOR Terms and Conditions
is only a small fraction of the factorizations and this fraction goes to zero as x → ∞.
So from Theorem 4.2 and Proposition 4.3 it follows that for sufficiently large x we get
about x/ log x factorizations which contain at most (x/M)/ log(x/M) ∼ x/M log(x)
distinct prime factors. So the number of factorizations exceeds M times the number of
prime factors. We now apply a slight generalization of Theorem 2.4 which works for
all k, rather than just prime power values. It can be found [1, Theorem 2] and it reads
as follows.
Theorem 4.4. If k ≥ 3 and r are positive integers, then any subset of (Z/kZ)r of
cardinality > r k log k contains a non-empty subset whose sum of elements is 0 mod k.
We apply this theorem to the set of exponent vectors modulo k of our factorizations.
These vectors have length about x/(M log x) and they are x/ log x in number. Hence,
as we remarked before, this cardinality exceeds M times the length of the exponent
vectors. Since M > k log k, Theorem 4.4 can be invoked and we find a non-trivial
relation, and hence a number whose divisor sum is a kth power. By choosing x as
large as we like, we can produce infinitely many examples in this way, and Theorem
2.2 is proven.
REFERENCES
1. W. R. Alford, A. Granville, C. Pomerance, There are infinitely many Carmichael numbers, Ann. of Math.
139 (1994) 703–722; available at http://dx.doi.org/10.2307/2118576.
2. W. D. Banks, J. B. Friedlander, C. Pomerance, I. E. Shparlinski, Multiplicative structure of values of the
Euler function, in High Primes and Misdemeanours: Lectures in Honour of the 60th Birthday of Hugh
Cowie Williams, Edited by A. van der Poorten, A. Stein, and H. Williams, Fields Inst. Commun., Vol. 41,
Amer. Math. Soc., Providence, RI, 2004. 29–47.
3. L. E. Dickson, History of the Theory of Numbers, Vol. I, Divisibility and Primality, Chelsea Publishing,
New York, 1966.
4. P. van Emde Boas, D. Kruyswijk, A combinatorial problem on finite Abelian groups, Math. Centrum
Amsterdam Afd. Zuivere Wisk. 1967 (1969) ZW-009.
5. T. Freiberg, Products of shifted primes simultaneously taking perfect power values (2010), available at
http://arxiv.org/abs/1008.1978.
6. H. Halberstam, H.-E. Richert, Sieve Methods, London Mathematical Society Monographs, No. 4. Academic Press, London-New York, 1974.
7. G. H. Hardy, E. M. Wright, An Introduction to the Theory of Numbers, fifth edition. The Clarendon Press,
Oxford University Press, New York, 1979.
8. T. E. Mason, Problems and Solutions, Amer. Math. Monthly 23 (1916) 394; available at http://dx.
doi.org/10.2307/2971845.
9. Problem section, Nieuw Archief voor Wiskunde 11 5th series (2010) 149.
10. J. E. Olson, A combinatorial problem on finite Abelian groups I, II, J. Number Theory 1 (1969) 8–
11 available at http://dx.doi.org/10.1016/0022-314X(69)90021-3; and 195–199 available at
http://dx.doi.org/10.1016/0022-314X(69)90037-7.
11. N. J. Sloane, The Online Encyclopedia of integer sequences, available at http://oeis.org.
12. D. B. Zagier, Newman’s short proof of the prime number theorem, Amer. Math. Monthly 104 (1997)
705–708; available at http://dx.doi.org/10.2307/2975232.
FRITS BEUKERS received his Ph.D. in 1979 at the University of Leiden, Netherlands as a student of Rob
Tijdeman. After a one year membership at the Institute for Advanced Study, Princeton, he became an assistant
professor in Leiden. He is currently working as full professor at the University of Utrecht. His main research
interests range from number theory to arithmetic of linear differential equations, and hypergeometric functions.
Departement of Mathematics, University of Utrecht
[email protected]
FLORIAN LUCA received a B.S. in Mathematics from the University of Iasi, Romania in 1991 and a Ph.D.
in Mathematics from the University of Alaska, Fairbanks in 1996. After a few visiting positions at Syracuse
May 2012]
POWER VALUES OF DIVISOR SUMS
This content downloaded on Fri, 15 Feb 2013 04:05:44 AM
All use subject to JSTOR Terms and Conditions
379
University, Bielefeld University in Germany, and the Czech Academy of Sciences in Prague, he accepted an
appointment at the Universidad Nacional Autonóma de México in Morelia, Mexico where he is still today. His
research interests revolve around questions from elementary number theory and diophantine equations.
Instituto de Matemáticas, Universidad Nacional Autónoma de México, Morelia
[email protected]
FRANS OORT did his Ph.D. research mainly in Pisa (under Aldo Andreotti) and in Paris (under Jean-Pierre
Serre), and he obtained his Ph.D. degree in 1961 at Leiden University. He has held positions in Amsterdam
and in Utrecht, and he has held many visiting positions (such as Harvard, Aarhus, IAS, Kyoto University, MIT,
and Columbia University). Although retired since 2000, he is still actively engaged in research in algebraic
geometry, more specifically in moduli spaces of abelian varieties in positive characteristic.
Departement of Mathematics, University of Utrecht
[email protected]
Irrational Numbers Redux
In a 1973 M ONTHLY Classroom Note [1] called “Irrational Numbers”, the following non-constructive short argument was given to show that an irrational
number
raised to an irrational
power may be rational. Consider the identity
√ √2
√ √2 √2
√ √2
( 2 ) = 2. If 2√ is rational, then we are done. If not, then 2 is
√ 2 √
irrational, and so ( 2 ) 2 serves as an example. It was also observed there
along similar lines that an irrational number
raised to an irrational power may be
√ √2+1 √ √2 √
irrational, by considering the identity 2
2.
= 2
These proofs, while being elementary, are non-constructive in that the logical
principle of the excluded middle (tertium non datur) is used,
which the intuition√ √2
ists reject. As remarked also in [1], it √
is known that 2 is irrational, since it is
the square root of Hilbert’s number 2 2 , which was proved to be transcendental
by Kuzmin [2] in 1930. However, this result is not elementary, and is not used
above.
In this note we give a constructive elementary proof of the fact that an irrational number raised to an irrational power can be rational.
First we note that log2 3 is irrational, for otherwise if log2 3 = qp , where p and
q are positive integers, then 2 p = 3q , which is impossible, as 2 p is even while 3q
√ (2log2 3)
is odd. Our concrete example is then given by 2
= 3, which is rational,
√
√ log2 3 √
and 2 and 2log2 3 are irrational. Also, 2
= 3 shows that an irrational
number raised to an irrational power may be irrational.
REFERENCES
1. J.P. Jones and S. Toporowski. Irrational numbers. Amer. Math. Monthly, 80 (1973) 423–424.
2. R. Kuzmin. On a new class of transcendental numbers. Izv. Akad. Nauk SSSR, Ser. Mat., 7 (1930)
585–597.
—Submitted by AMOL SASANE, Stockholm, Swenden
380
c THE MATHEMATICAL ASSOCIATION OF AMERICA [Monthly 119
This content downloaded on Fri, 15 Feb 2013 04:05:44 AM
All use subject to JSTOR Terms and Conditions